1. ## Integration help

So is it safe to say that we can use integration by substitution???

2. ## Re: Integration help

I would simplify the integrand by factoring $\sqrt{t}$ from the numerator, then dividing out the common factors of the numerator and denominator. Or you might find it easier to just split the fraction, then reduce each term.

3. ## Re: Integration help

carry out the division

4. ## Re: Integration help

Can' i just put √t on the numerator and make it t^-1/2 and multiply (t + 2t^2)t^-1/2 together and integrate????

5. ## Re: Integration help

Can' i just put √t on the numerator and make it t^-1/2 and multiply (t + 2t^2)t^-1/2 together and integrate????.....

6. ## Re: Integration help

Yes, that's another way of splitting the fraction.

7. ## Re: Integration help

yup.

you'd end up with $\frac{t}{t^{1/2}}+\frac{2t^2}{t^{1/2}}=t^{1/2}+2*t^{3/2}$

or $\int t^{1/2}+2*t^{3/2}dx$

8. ## Re: Integration help

then im authorized to integrate after that step right?

9. ## Re: Integration help

Yes, now it is just a matter of applying the power rule term by term.

10. ## Re: Integration help

Originally Posted by asilvester635
So is it safe to say that we can use integration by substitution???
You can do this with the help of integration by parts.

Now after simplification:

$\int \frac{t + 2t^2}{\sqrt{t}}\;\; dt = \int (\sqrt{t} + 2t\sqrt{t})\;\;dt = \int \sqrt{t}\;\;dt + \int 2t\sqrt{t}\;\;dt$

We divide this integration in two parts. First:

$\int \sqrt{t}\;\;dt = \frac{2}{3}t^{\frac{3}{2}} + C_1$

Then for $\int 2t\sqrt{t}\;\; dt$ we use integration by parts:

$\text{Let } u = t \;\;\therefore du = dt$

$\text{Let } dv = 2\sqrt{t}\;\; dt\;\; \therefore v = \frac{4}{3}t^{\frac{3}{2}}$

Now:

\begin{align*}\int 2t\sqrt{t}\;\;dt =& \frac{4}{3}t^{\frac{5}{2}} - \int \frac{4}{3}t^{\frac{3}{2}}\;\; dt \\ =& \frac{4}{3}t^{\frac{5}{2}} - \frac{8}{15}t^{\frac{5}{2}} + C_2 \\ =& \frac{4}{5}t^{\frac{5}{2}}+ C_2...........\text{[After simplicfication]}\end{align*}

So:

$\therefore \int \frac{t + 2t^2}{\sqrt{t}}\;\; dt = \frac{2}{3}t^{\frac{3}{2}} + \frac{4}{5}t^{\frac{5}{2}} + C$

Check:

integration (t + 2*t^2)/(t)^(1/2) dt - Wolfram|Alpha